Question

Electron accelerated by potential $V$ are diffracted from a crystal. If $d=1A$ and $i={30}^{\circ }$. $V$ should be about $h=6.6\times {10}^{-24}Js{m}_e=9.1\times {10}^{-33}kg.e=1.6\times {10}^{-19}C$

• A. $2000V$
• B. $50V$
• C. $500V$
• D. $1000V$

Answer 1

The correct option is B $50V$

$d=1A$

$i={30}^o$

$\theta ={60}^o$

$h=6.6\times {10}^{-34}Js$

$m_e=9.1\times {10}^{-31}kg$

$e=1.6\times {10}^{-19}c$

$n\lambda =2dsin\theta$

$\lambda =\frac{2\times 1A\times sin{60}^o}{1}$

$\lambda =\sqrt{3A}$

$\sqrt{V}=\frac{1.27\times {10}^{-10}}{\sqrt{3}\times {10}^{-10}}=50.18volts$